Problem: Simplify and expand the following expression: $ \dfrac{4}{p + 2}- \dfrac{4}{4p + 8}- \dfrac{2p}{p^2 + 4p + 4} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{4}{4p + 8} = \dfrac{4}{4(p + 2)}$ We can factor the quadratic in the third term: $ \dfrac{2p}{p^2 + 4p + 4} = \dfrac{2p}{(p + 2)(p + 2)}$ Now we have: $ \dfrac{4}{p + 2}- \dfrac{4}{4(p + 2)}- \dfrac{2p}{(p + 2)(p + 2)} $ The least common multiple of the denominators is: $ (p + 2)(p + 2)$ In order to get the first term over $(p + 2)(p + 2)$ , multiply by $\dfrac{4(p + 2)}{4(p + 2)}$ $ \dfrac{4}{p + 2} \times \dfrac{4(p + 2)}{4(p + 2)} = \dfrac{16(p + 2)}{(p + 2)(p + 2)} $ In order to get the second term over $(p + 2)(p + 2)$ , multiply by $\dfrac{p + 2}{p + 2}$ $ \dfrac{4}{4(p + 2)} \times \dfrac{p + 2}{p + 2} = \dfrac{4(p + 2)}{(p + 2)(p + 2)} $ In order to get the third term over $(p + 2)(p + 2)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{2p}{(p + 2)(p + 2)} \times \dfrac{4}{4} = \dfrac{8p}{(p + 2)(p + 2)} $ Now we have: $ \dfrac{16(p + 2)}{(p + 2)(p + 2)} - \dfrac{4(p + 2)}{(p + 2)(p + 2)} - \dfrac{8p}{(p + 2)(p + 2)} $ $ = \dfrac{ 16(p + 2) - 4(p + 2) - 8p} {(p + 2)(p + 2)} $ Expand: $ = \dfrac{16p + 32 - 4p - 8 - 8p}{4p^2 + 16p + 16} $ $ = \dfrac{4p + 24}{4p^2 + 16p + 16}$ Simplify: $ = \dfrac{p + 6}{p^2 + 4p + 4}$